3. Finding a formula for the inverse of a function can be extremely hard even if we know the inverse exists. Miraculously, if the function is analytic then its inverse can be explicitly computed as a power series: Theorem. Let I be an open interval and let f be an analytic function on I. Fix a E I and l) = a). IF V5: E I, ftp") y 0 THEN f'l exists and \"is also analytic on its domain. Moreover, fmr y near b, 00 Flo) = Z 309 - 5)\" n:(l where on = f'1(l7) = a and n_1 \"Pa n v... e = a. [in1 HM) ll You will assume1 this theorem to compute the inverse of f(.r) : item as power series. (Try the usual approach to nding an inverse of f. Youjll quickly see it's impossible.) (a) Find the largest open interval I centered at 0 such that f is one-to-one on I. (b) Explain why f is analytic on I. The explanation should be short. (C) Let f"1 be the inverse of f on I. Compute f'1 as a power series expansion centered at 0 using the above theorem. Your answer should be a series written using sigma notation. \f(f) Let a: E I be the value such that 3351' = i. Approximate 53 Within an error less than 0.01. Your answer should be expressed as a fraction. Note that Z is in the interval convergence for the power series from (3c)